JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    As per this diagram a point charge \[+q\] is placed at the origin \[O\]. Work done in taking another point charge \[-Q\] from the point \[A\] [co-ordinates \[(0,\,a)\]] to another point B [co-ordinates (a, 0)] along the straight path \[AB\] is [CBSE PMT 2005]

    A)            Zero

    B)            \[\left( \frac{-qQ}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}} \right)\,\sqrt{2}a\]

    C)            \[\left( \frac{qQ}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}} \right)\,\frac{a}{\sqrt{2}}\]

    D)            \[\left( \frac{qQ}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}} \right)\,\sqrt{2}a\]

    Correct Answer: A

    Solution :

                 Since A and B are at equal potential so potential difference between A and B is zero. Hence \[W=Q.\Delta V=\text{ }0\]


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